The functoriality of the cellular homology Given $\pi：S^n \to \mathbb RP^n$ the usual quotient map, I am trying to figure out what is the induced map on the $nth$ homology. 
If I use the singular homology it seems hopeless, since it is hard to locate an explicit generator. So I decide to use the CW complex structure on $S^n$ which consists of 1 $n$ cell and 1 $0$ cell. Intuitively speaking, the $1$ cell is mapped to the $1$ cell in $\mathbb RP^n$ "twice". But how do you make sense of this rigorously?
Update:
The question I really have in mind is when we look at the induced map $\pi_{\ast}$ between cellular homology, by definition, we are looking at the induced map between relative homology. However, we are making the identification such that $H_n(S^n, S^{n-1})$ is free abelian with basis its n-cells. Using this identification how should I calculate $\pi_{\ast}$? More specifically, what does it correspond to as a map between cells?
There is a related question: 
Morphism induced by a cellular map between CW-complexes
 A: The map is trivial when n is even and I can show that it is either multiplication by 1 or 2. This is not a full answer (Edit: I made a mistake. It follows that it is 2 from what I said). Note that this is easy to see when n is even since $H_n(\mathbb{R}P^n)=0$.
Now in the case n is odd. Use the cell structure for $S^n$ with 1 0-cell and 1 n-cell. Then the nth chain group for its cellular chain complex is just its usual nth homology of the n-sphere. The induced map on the chain level or homology level (since the image of the boundary map is trivial and the kernel is the entire group) is the map $H_n(S^n) \rightarrow H_n(\mathbb{R}P^n) \rightarrow H_n(S^n)$ where the first map is the map in question and the last map is collapsing the (n-1)-skeleton. This composition has degree 2 when n is odd (it is the same computation as the differential for the cellular chain complex of $\mathbb{R}P^n$). This tells you that the map $H_n(S^n) \rightarrow H_n(\mathbb{R}P^2)$ is multiplication by 2 on the chain level and on the homology level.
If you want to check that this calculation is right, it is simple to consider the $n=1$ case where the map $S^1 \rightarrow \mathbb{R}P^1=S^1$ is squaring which is multiplication by 2 on the homology level.
Edit: Actually, it's written right in what I said. It is multiplication by 2. Writing it as a composition is unnecessary, but it does tell you that the quotient map $\mathbb{R}P^n \rightarrow S^n$ induces isomorphism on the nth homology if n is odd.
